Today's lesson
So far we've used trig to find sides and angles when a triangle is already drawn for us. Today's job is to read a word problem, draw your own right-angled triangle, and use the same SOHCAHTOA tools to answer the question.
Learning intentions
- Know how the angle of elevation is measured up from the horizontal
- Know how the angle of depression is measured down from the horizontal
- Read a word problem and draw a labelled right-angled triangle
- Pick the right trig ratio and answer the question in a sentence
Part 1 — Elevation vs depression (~6 min)
Two angles, measured from the horizontal
- Angle of elevation — looking up from horizontal to a point above you (e.g. looking at a kite, a helicopter, the top of a tower).
- Angle of depression — looking down from horizontal to a point below you (e.g. lighthouse keeper looking down at a boat).
- On the same diagram, elevation and depression between two points are equal — they are alternate angles in parallel lines (one horizontal at the top, one at the bottom).
- Draw a clean diagram with the horizontal ground line and the vertical object.
- Mark the angle from the horizontal (elevation up, depression down).
- Label what you know and put a letter on what you want.
- Decide which sides you know/want → pick sin, cos, or tan.
- Solve, then write the answer in a full sentence with units.
Part 2 — Angle of elevation (Example 5, ~12 min)
📺 Walkthrough: a helicopter hovers 250 m up; the angle of elevation from the helipad is 35°. Find the horizontal distance.
- Draw the right triangle: 250 m is vertical (opposite the 35° angle), $x$ m is horizontal (adjacent), and the slant is the line of sight.
- $x$ is adjacent, $250$ is opposite → use $\tan$.
- $\tan 35^\circ=\dfrac{250}{x}$
- $x\tan 35^\circ=250$
- $x=\dfrac{250}{\tan 35^\circ}=\boxed{357.04\text{ m}}$ (to the nearest cm)
- The horizontal distance from the helipad to the helicopter is 357.04 m.
Now you try: A bird is hovering at 100 m. The angle of elevation from an observation point to the bird is 55°. Find the horizontal distance. Answer: $x=\dfrac{100}{\tan 55^\circ}\approx 70.02$ m.
Practice 2.1 — elevation problems (2 d.p.)
- A kite is flying directly above a point on the ground. The string is 80 m long and makes an angle of elevation of 62° with the ground. How high is the kite?
- The angle of elevation from a point on the ground to the top of a 45 m tall flagpole is 38°. How far is the point from the base of the pole?
- A 45 m guy-wire is anchored from the ground to the top of a communications mast at an elevation of 60°. How tall is the mast?
b) distance = $\dfrac{45}{\tan 38^\circ}\approx 57.59$ m
c) mast = $45\sin 60^\circ\approx 38.97$ m
Part 3 — Angle of depression (Example 6, ~12 min)
📺 Walkthrough: two buildings 57 m apart, 237 m and 158 m tall — find the angle of depression from the taller to the shorter.
- Draw a horizontal line at the top of the taller building, and the line of sight down to the top of the shorter.
- Height difference $=237-158=79$ m. This is the opposite side of the right triangle (vertical drop).
- The horizontal gap $57$ m is the adjacent side.
- $\tan\theta=\dfrac{79}{57}$
- $\theta=\tan^{-1}\!\left(\dfrac{79}{57}\right)=\boxed{54.19^\circ}$ (to 2 d.p.)
- The angle of depression is 54.19°.
Now you try: Two vertical poles 32 m apart are 62 m and 79 m high. Find the angle of depression from the top of the taller pole to the top of the shorter. Answer: $\theta=\tan^{-1}\!\left(\dfrac{17}{32}\right)\approx 27.98°$.
- Convert 1.5 km to 1500 m so units match.
- Vertical drop (opposite) = 112 m; horizontal (adjacent) = 1500 m.
- $\tan\theta=\dfrac{112}{1500}$
- $\theta=\tan^{-1}\!\left(\dfrac{112}{1500}\right)=\boxed{4.27°}$ (to 2 d.p.)
Practice 3.1 — depression problems (2 d.p.)
- From the top of a 25 m tall viewing tower, the angle of depression to a crocodile on the ground is 62°. Find the direct (slant) distance from the top of the tower to the crocodile.
- Two towers, 80 m apart, are 30 m and 50 m tall. Find the angle of depression from the top of the taller tower to the top of the shorter tower.
- A drone is 60 m above a flat oval. The angle of depression from the drone to a player on the ground is 18°. How far away (horizontally) is the player from a point directly below the drone?
b) $\theta=\tan^{-1}\!\left(\dfrac{20}{80}\right)\approx 14.04°$
c) distance = $\dfrac{60}{\tan 18^\circ}\approx 184.66$ m
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The angle of elevation is measured…
Q2. On the same diagram with two horizontal lines, the angle of elevation from A to B is equal to the angle of depression from B to A because they are…
Q3. A helicopter is 200 m up. The angle of elevation from the ground is 40°. Horizontal distance is found by:
Q4. From a 40 m cliff, the angle of depression to a boat is 25°. The horizontal distance from the cliff base to the boat (2 d.p.) is:
Q5. Two trees 12 m apart on flat ground are 8 m and 14 m tall. The angle of depression from the top of the taller tree to the top of the shorter is (2 d.p.):
Working program — Cambridge Ex 6C
After the quiz, open Cambridge Chapter 6 and complete the following:
| Section | Foundation | Standard (set work) | Advanced |
|---|---|---|---|
| Fluency 1–6 | 1–5 | 1–5 | 2, 4–6 |
| Problem-solving 7–9 | 7 | 7, 8 | 8, 9 |
| Reasoning 10–13 | 10 | 10, 11 | 11–13 |
| Enrichment 14 | — | — | 14 |
If you finish early, preview Ex 6D — Directions and bearings (tomorrow's lesson).
Exit ticket — write in your book
- From which line (horizontal or vertical) is the angle of elevation measured?
- If the angle of elevation from A to B is 40°, what is the angle of depression from B to A? Why?
- You're at the top of a 30 m tower and see a person at a depression of 20°. Which trig ratio gives you their horizontal distance from the base?